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my question is a general one, can a unit vector after transformation have the length longer than length of the eigenvector which highest eigenvalue? is there any proof?

purpose is that I want to know in the principal component analysis is the eigenvector which highest eigenvalue is the direction with widest data variance in that direction or it's just better option or even calcutable.

Farhang Amaji
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    Generally speaking, the answer is no. For example, $$ T = \pmatrix{1&1\\0&0} $$ has eigenvalues $0,1$, but with $v = \pmatrix{1 & 1}$ we find that $\|Tv\| = \sqrt{2}\|v\|$. – Ben Grossmann Sep 13 '21 at 01:46
  • A correct remark, thanks to @BenGrossman for pointing out a silly blunder in my previous: for _self-adjoint_ operator $T$, the length of $Tv$ is at most $|\lambda|$ times the length of $v$, where $\lambda$ is the largest eigenvector. For not-self-adjoint operators, even diagonalizable, this is false in general. So, after this correction, is this the sort of thing you are thinking in terms of? – paul garrett Sep 13 '21 at 01:52
  • @paul To add to that, the same can be said for [normal operators](https://en.wikipedia.org/wiki/Normal_operator) – Ben Grossmann Sep 13 '21 at 01:53
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    @Farhang In the context of PCA, one is typically concerned with the (symmetric!) matrix $M = X^TX$, where $X$ is a data matrix (normalized to have column mean zero). For such a matrix, the largest eigenvalue is indeed the largest factor by which a vector's length can be increased under multiplication by $M$. – Ben Grossmann Sep 13 '21 at 01:56
  • Related: https://math.stackexchange.com/questions/1437569/do-eigenvectors-correspond-to-direction-of-maximum-scaling – Theo Bendit Sep 13 '21 at 02:12

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